Answer :
To find the volume of a gas when it undergoes changes in pressure and temperature, we can use the Combined Gas Law. This law combines Boyle's Law, Charles's Law, and Gay-Lussac's Law. It relates the pressure (P), volume (V), and temperature (T) of a gas in two different states.
The formula for the Combined Gas Law is:
[tex]\[
\frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}
\][/tex]
Where:
- [tex]\(P_1\)[/tex], [tex]\(V_1\)[/tex], and [tex]\(T_1\)[/tex] are the initial pressure, volume, and temperature.
- [tex]\(P_2\)[/tex], [tex]\(V_2\)[/tex], and [tex]\(T_2\)[/tex] are the final pressure, volume, and temperature.
Given:
- Initial volume, [tex]\(V_1 = 115 \, \text{mL}\)[/tex]
- Initial pressure, [tex]\(P_1 = 45.8 \, \text{kPa}\)[/tex]
- Initial temperature, [tex]\(T_1 = 57.2^\circ \text{C}\)[/tex]
- Final pressure, [tex]\(P_2 = 99.3 \, \text{kPa}\)[/tex]
- Final temperature, [tex]\(T_2 = 12.3^\circ \text{C}\)[/tex]
Step 1: Convert the temperatures from Celsius to Kelvin.
To convert from Celsius to Kelvin, use the formula:
[tex]\[ T(\text{K}) = T(\text{C}) + 273.15 \][/tex]
So,
- [tex]\(T_1 = 57.2 + 273.15 = 330.35 \, \text{K}\)[/tex]
- [tex]\(T_2 = 12.3 + 273.15 = 285.45 \, \text{K}\)[/tex]
Step 2: Use the Combined Gas Law to solve for the final volume, [tex]\(V_2\)[/tex].
Rearrange the formula to solve for [tex]\(V_2\)[/tex]:
[tex]\[
V_2 = \frac{P_1 \times V_1 \times T_2}{P_2 \times T_1}
\][/tex]
Plug in the values:
[tex]\[
V_2 = \frac{45.8 \times 115 \times 285.45}{99.3 \times 330.35}
\][/tex]
Simplify and calculate:
[tex]\[
V_2 \approx 45.83 \, \text{mL}
\][/tex]
So, the volume of the gas at a pressure of 99.3 kPa and a temperature of [tex]\(12.3^\circ \text{C}\)[/tex] is approximately 45.83 mL.
The formula for the Combined Gas Law is:
[tex]\[
\frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}
\][/tex]
Where:
- [tex]\(P_1\)[/tex], [tex]\(V_1\)[/tex], and [tex]\(T_1\)[/tex] are the initial pressure, volume, and temperature.
- [tex]\(P_2\)[/tex], [tex]\(V_2\)[/tex], and [tex]\(T_2\)[/tex] are the final pressure, volume, and temperature.
Given:
- Initial volume, [tex]\(V_1 = 115 \, \text{mL}\)[/tex]
- Initial pressure, [tex]\(P_1 = 45.8 \, \text{kPa}\)[/tex]
- Initial temperature, [tex]\(T_1 = 57.2^\circ \text{C}\)[/tex]
- Final pressure, [tex]\(P_2 = 99.3 \, \text{kPa}\)[/tex]
- Final temperature, [tex]\(T_2 = 12.3^\circ \text{C}\)[/tex]
Step 1: Convert the temperatures from Celsius to Kelvin.
To convert from Celsius to Kelvin, use the formula:
[tex]\[ T(\text{K}) = T(\text{C}) + 273.15 \][/tex]
So,
- [tex]\(T_1 = 57.2 + 273.15 = 330.35 \, \text{K}\)[/tex]
- [tex]\(T_2 = 12.3 + 273.15 = 285.45 \, \text{K}\)[/tex]
Step 2: Use the Combined Gas Law to solve for the final volume, [tex]\(V_2\)[/tex].
Rearrange the formula to solve for [tex]\(V_2\)[/tex]:
[tex]\[
V_2 = \frac{P_1 \times V_1 \times T_2}{P_2 \times T_1}
\][/tex]
Plug in the values:
[tex]\[
V_2 = \frac{45.8 \times 115 \times 285.45}{99.3 \times 330.35}
\][/tex]
Simplify and calculate:
[tex]\[
V_2 \approx 45.83 \, \text{mL}
\][/tex]
So, the volume of the gas at a pressure of 99.3 kPa and a temperature of [tex]\(12.3^\circ \text{C}\)[/tex] is approximately 45.83 mL.
